\documentclass[windows,csize4]{BHCexam}
%\documentclass[windows,csize4,answers]{BHCexam}

\usepackage{multicol} % 分栏
\pagestyle{fancy}
\fancyfoot[C]{\kaishu \small 第 \thepage 页 共 \pageref{lastpage} 页}
%\fancyhead[L]{\includegraphics[width=2cm]{qrcode.png}}
\title{因式分解 - 双十字相乘，主元法作业}
%\subtitle{数学文科试卷}
%\notice{满分150分, 120分钟完成, \\	允许使用计算器，答案一律写在答题纸上.}
%\author{Gavin Chen}
%\date{\today}
\usepackage{enumerate} % 编号

\begin{document}
\maketitle


\begin{groups}

    \group{用双十字相乘因式分解}{}
    \begin{questions}[]

        \begin{multicols}{2}
            \question[5] $x^2-3xy-10y^2+x+9y-2$
            \begin{solution}{0.5cm}
                \methodonly $na$
            \end{solution}

            \question[5] $x^2+2xy-3y^2+2xz+14yz-8z^2$
            \begin{solution}{0.5cm}
                \methodonly
            \end{solution}
        \end{multicols}
        \vspace{3.5cm}

        \begin{multicols}{2}
            \question[5] $2x^2+xy-y^2-4x+5y-6$
            \begin{solution}{0.5cm}
                \methodonly
            \end{solution}

            \question[5] $x^2+x-4 y^2+10 y-6$
            \begin{solution}{0.5cm}
                \methodonly $(x-2 y+3) (x+2 y-2)$
            \end{solution}
        \end{multicols}
        \vspace{3.5cm}

        \begin{multicols}{2}
            \question[5] $3x^2+5xy-2y^2+x+9y-4$
            \begin{solution}{0.5cm}
                \methodonly
            \end{solution}

            \question[5] $x^2+3xy+2y^2+4x+5y+3$
            \begin{solution}{0.5cm}
                \methodonly
            \end{solution}
        \end{multicols}
        \vspace{3.5cm}
    \end{questions}

    \group{用主元法因式分解}{}
    \begin{questions}[]
            \question[5] $2x^2-7xy+3y^2+5xz-5yz+2z^2$
            \begin{solution}{0.5cm}
                \methodonly $(2x-y+z)(x-3y+2z)$
            \end{solution}
            \vspace{3.5cm}

            \question[5] $x^3+(2y+1)x^2+(y^2+2y-1)x+y^2-1$
            \begin{solution}{0.5cm}
                \methodonly 此题用$x$做主元无法分解，故而可选取$y$做主元。
                \[
                    \begin{aligned}
                         & \phantom{=}x^3+(2y+1)x^2+(y^2+2y-1)x+y^2-1 \\
                         & =x^3+2x^2y+x^2+xy^2+2xy-x+y^2-1        \\
                         & =(x+1)y^2+(2x^2+2x)y+(x^3+x^2-x-1) \\
                         & =(x+1)\left[y^2+2xy+(x^2-1)\right] \\ 
                         & =(x+1)(x+y+1)(x+y-1)
                    \end{aligned}
                \]
            \end{solution}
        \vspace{3.5cm}

        \question[5] $x^2y-xy^2+x^2z-xz^2-3xyz+y^2z+yz^2$
        \begin{solution}{0.5cm}
            \methodonly 此题用$x$做主元
            \[
                \begin{aligned}
                     & \phantom{=}x^2y-xy^2+x^2z-xz^2-3xyz+y^2z+yz^2 \\
                     & =(y+z)x^2-(y^2+z^2+3yz)x+yz(y+z)  \\
                     & =\left[(y+z)x-yz\right]\left[x-(y+z)\right] \\ 
                     & = (xy+xz-yz)(x-y-z)
                \end{aligned}
            \]
        \end{solution}
        \vspace{3.5cm}

        \question[5] $9-3a+3b+3ab-a^2b+ab^2$
        \begin{solution}{0.5cm}
            \methodonly 此题可以用$a$做主元然后十字相乘，但用常数$3$做主元更方便
            \[
                \begin{aligned}
                     & \phantom{=}9-3a+3b+3ab-a^2b+ab^2 \\
                     & =(3+ab)\left[3-(a-b)\right]  \\ 
                     & =(3+ab)(3-a+b)
                \end{aligned}
            \]
        \end{solution}

    \end{questions}

\end{groups}






\label{lastpage}
\end{document}